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It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this. I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant ...


9

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the ...


8

The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time. $$ \begin{array}{lll} \hline \text{Reference} & R_\ce{O} & R_\ce{N} \\ \hline \text{Pauling, 1939} & 1.40 & 1.5 \\ \text{Bondi, 1964} & 1.52 & 1.55 \\ \text{Zefirov, 1974} & 1.29 & 1.50 \\ \text{Gavezzotti,...


4

There are a few ways that electronegativity can be measured/calculated, the original being that of Pauling, followed by others scales such as the Allred-Rochow and Mulliken electronegativity. When looking at the values obtained by Pauling, one must keep in mind that these values aren't grounded in quantum mechanics (i.e. electronegativity does not have an ...


4

The fallacy is the assumption that "inner $d$ orbitals" become "available". Generally they are not, in the Group 1 and Group 2 metals. There are rare occasions where some evidence of $d$-orbital bonding is found for heavier G2 elements, but the impact is small and not widespread. In Groups 1 and 2, where there are inner $d$ orbitals ...


3

The bond with greater electronegativity difference has greater tendency to break with help of polar $\ce{H2O}$ molecules to hydrated ions, unless it is ionic already. Electrons then stay at the oxygen as the most electronegative atom from all the three. So if $\ce{M}$ is less electronegative then $\ce{H}$, there will be' $$\ce{M^{(+)}-O^{(-)}-H(aq) <=> ...


3

Slater's rules are an attempt to lump the effect of all other electrons on the wavefunction, and thereby other properties such as energy, of an electron (described by a hydrogen-like wavefunction). The effect of electron-electron repulsion is modeled indirectly by saying that inner shell electrons effectively screen the attractive nuclear charge sensed by ...


2

Firstly, we need to clearly define "atomic volume" and "molar volume". Typically, one would understand "atomic volume" to be the volume of an atom of the element and it is directly related to the radius of the atom. Molar volume is a macroscopic property that is defined by Wikipedia as the ratio of the molar mass to the mass ...


2

The Ionisation Energy of an atom is defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. In this case, since $E_1$ amount of energy is used to ionise $N_0/2$ atoms, the ionisation energy of each atom is $\frac{E_1}{N_0/2} = \frac{2E_1}{N_0}$. The Electron Affinity is ...


2

Put very roughly, we can boil this down to why aluminum and sulfur are depressed with respect to the average of the surrounding elements. Aluminum is $\ce{[Ne]{3}s^2{3}p^1}$. Its first ionization removes the $3p$ electron while leaving the more stable $3s$ orbital intact, whereas magnesium has to break its more stable $3s$ orbital. So by comparison, the ...


2

As you surmise, quantum mechanics was developed to help explain observations such as that periodicity, spectral lines and the photoelectric effect. However, the solution of the Schrödinger equation is difficult, and not complete beyond hydrogen. For example Wikipedia states, "Unlike for hydrogen, a closed-form solution to the Schrödinger equation for ...


1

You expect ionization energies to increase as you remove more electrons from an atom, because the charge of the nucleus remains the same (and therefore its attraction to the remaining electrons) but repulsion from other electrons is reduced with removal of each additional electron. However, in addition to this simple intuitive trend, quantum mechanics throws ...


1

There is the general trend in the groups 15, 16, 17 of raising of boiling points for the binary compounds with hydrogen, going down the groups. But the first members of each group - $\ce{NH3, H2O, HF}$ - have anomally with their boiling points being exceptionally high, due hydrogen bonds. The strength of these bonds and the boiling point decreases in the ...


1

Since I am neither clairvoyant nor able to travel back and forth in time, I can only guess what IUPAC was thinking. But there is no guessing that the inner transition elements do not completely fit in with Group 3 chemically. One major chemical difference is in the oxidation states the elements may achieve, and thus the stoichiometry of compounds with ...


1

The reason is that after arsenic, there is shielding. This causes the intermolecular forces to become weak. Thus, the melting point decreases. I hope you know about shielding of d and f electrons


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